Explanation

(b) Let $R$ be the resistance of each bulb.
We have $P = \frac{V^{2}}{R}$
or $R = \frac{V^{2}}{P} = \frac{V^{2}}{100}$
When the bulbs are connected in series, the voltage across each of them is $V/2$. Hence, the total power consumed is
$\ P^{'} = \frac{\left( \ \frac{V}{2} \right)^{2}}{R} + \frac{\left( \ \frac{V}{2} \right)^{2}}{R} = \frac{V^{2}}{2R}$
${\therefore P}^{'} = \frac{P}{2} = \frac{100}{2} =$50 W